Table of Contents
How to Use This Page
Read each concept carefully, then check the formula, common mistake, and exam tip before moving to the next. This page completely covers Atoms for CBSE Class 12 Physics, bridging classical electromagnetism with quantum mechanics.
Key Concepts
Atomic Structure
Journey to the center of the atom
1
Distance of Closest Approach ($r_0$)
Formula
In Rutherford’s $\alpha$-scattering experiment, an $\alpha$-particle directed straight at the nucleus comes to rest when its entire initial Kinetic Energy converts to Electrostatic Potential Energy. It gives an estimate of nuclear size.
$$K = \frac{1}{4\pi\epsilon_0} \frac{(2e)(Ze)}{r_0} \implies r_0 = \frac{1}{4\pi\epsilon_0} \frac{2Ze^2}{K}$$
2
Impact Parameter ($b$)
Formula
The perpendicular distance of the initial velocity vector of the $\alpha$-particle from the central line of the nucleus. A smaller impact parameter leads to a larger scattering angle ($\theta$).
$$b = \frac{1}{4\pi\epsilon_0} \frac{Ze^2 \cot(\theta/2)}{K}$$
3
Bohr’s Quantization Postulate
Formula
Electrons can only orbit in specific, stable, non-radiating paths called stationary orbits. The angular momentum ($L$) of an electron in the $n$-th orbit is an integral multiple of $h/2\pi$.
$$L = mvr = \frac{nh}{2\pi} \quad (n = 1, 2, 3 \dots)$$
4
Radius of Bohr’s Orbit ($r_n$)
Formula
The radius of the $n$-th allowed orbit is directly proportional to the square of the principal quantum number ($n$) and inversely proportional to the atomic number ($Z$).
$$r_n = \frac{n^2 h^2 \epsilon_0}{\pi m Z e^2} \approx 0.529 \frac{n^2}{Z} \text{ \AA}$$
5
Velocity of Electron ($v_n$)
Formula
The speed of an electron in the $n$-th orbit. It is inversely proportional to $n$, meaning electrons in outer orbits travel slower than those closer to the nucleus.
$$v_n = \frac{e^2}{2\epsilon_0 h} \frac{Z}{n} = \left(\frac{c}{137}\right) \frac{Z}{n}$$
6
Energy Levels in Hydrogen-like Atoms
Formula
The total energy ($E_n$) of an electron in the $n$-th orbit is negative, indicating the electron is bound to the nucleus.
$$E_n = – \frac{me^4}{8\epsilon_0^2 h^2} \frac{Z^2}{n^2} \approx -13.6 \frac{Z^2}{n^2} \text{ eV}$$
7
Energy Ratio ($K, U, E$)
Formula
The Kinetic Energy ($K$), Potential Energy ($U$), and Total Energy ($E$) of an electron in a Bohr orbit share a strict mathematical relationship.
$$K = -E \quad | \quad U = 2E \implies |K| : |U| : |E| = 1 : 2 : 1$$
8
Rydberg Formula for Spectral Lines
Formula
When an electron jumps from a higher orbit ($n_i$) to a lower orbit ($n_f$), it emits a photon. The wavelength ($\lambda$) of this photon is given by the Rydberg equation.
$$\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_f^2} – \frac{1}{n_i^2} \right)$$
$\text{Where } R \approx 1.097 \times 10^7 \text{ m}^{-1}$
9
De Broglie’s Justification of Bohr’s Postulate
Formula
De Broglie proved that an electron orbit is only stable if it forms a perfect standing matter wave around the nucleus. The circumference of the orbit must equal an integer number of wavelengths.
$$2\pi r = n\lambda = n\left(\frac{h}{mv}\right) \implies mvr = \frac{nh}{2\pi}$$
Concept Deep Dive
01
The Hydrogen Spectrum Series
Decoding the barcodes of the universeWhen excited electrons fall back down, they drop into specific target orbits, emitting light in specific regions of the electromagnetic spectrum:
1. Lyman Series ($n_f = 1$): Drops from $n=2,3,4\dots$ to $1$. Emits Ultraviolet (UV).
2. Balmer Series ($n_f = 2$): Drops from $n=3,4,5\dots$ to $2$. Emits Visible Light.
3. Paschen Series ($n_f = 3$): Drops from $n=4,5,6\dots$ to $3$. Emits Infrared (IR).
4. Brackett Series ($n_f = 4$): Drops to $4$. Emits Infrared (IR).
5. Pfund Series ($n_f = 5$): Drops to $5$. Emits Infrared (IR).
Astronomers use these exact spectral “fingerprints” to determine what distant stars are made of!
1. Lyman Series ($n_f = 1$): Drops from $n=2,3,4\dots$ to $1$. Emits Ultraviolet (UV).
2. Balmer Series ($n_f = 2$): Drops from $n=3,4,5\dots$ to $2$. Emits Visible Light.
3. Paschen Series ($n_f = 3$): Drops from $n=4,5,6\dots$ to $3$. Emits Infrared (IR).
4. Brackett Series ($n_f = 4$): Drops to $4$. Emits Infrared (IR).
5. Pfund Series ($n_f = 5$): Drops to $5$. Emits Infrared (IR).
Astronomers use these exact spectral “fingerprints” to determine what distant stars are made of!
02
Why the Negative Sign in Total Energy?
Escaping the potential wellThe total energy of an electron in the ground state of Hydrogen is $-13.6 \text{ eV}$. The negative sign is not arbitrary—it means the electron is in a Bound State.
By convention, the potential energy of an electron completely free from the nucleus ($r = \infty$) is zero. Because the nucleus attracts the electron, work is done by the electrostatic field as the electron falls into orbit, making its energy drop below zero. To free the electron (Ionization Energy), you must supply exactly $+13.6 \text{ eV}$ to bring its total energy back up to $0$.
By convention, the potential energy of an electron completely free from the nucleus ($r = \infty$) is zero. Because the nucleus attracts the electron, work is done by the electrostatic field as the electron falls into orbit, making its energy drop below zero. To free the electron (Ionization Energy), you must supply exactly $+13.6 \text{ eV}$ to bring its total energy back up to $0$.
Compare & Contrast
✗ Rutherford’s Model
- Electrons can revolve in orbits of any radius.
- Cannot explain the stability of the atom (accelerating charge should radiate energy and spiral into the nucleus).
- Cannot explain discrete (line) emission spectra; predicts a continuous spectrum.
- Based purely on classical physics.
✓ Bohr’s Model
- Electrons revolve ONLY in specific discrete (quantized) orbits.
- Explains stability: Electrons in these stationary orbits do not radiate energy.
- Perfectly explains the discrete line spectrum of Hydrogen.
- A hybrid of classical mechanics and early quantum mechanics.
Common Mistakes to Avoid
Mistake 1
Shortest vs. Longest Wavelength: In the Rydberg formula, energy and wavelength are inversely proportional ($E = hc/\lambda$).
– Shortest $\lambda$ (Max Energy): Electron drops from $n_i = \infty$ (Series Limit).
– Longest $\lambda$ (Min Energy): Electron drops from the very next immediate orbit (e.g., $n=3 \rightarrow n=2$ for Balmer).
– Shortest $\lambda$ (Max Energy): Electron drops from $n_i = \infty$ (Series Limit).
– Longest $\lambda$ (Min Energy): Electron drops from the very next immediate orbit (e.g., $n=3 \rightarrow n=2$ for Balmer).
Mistake 2
Excited State numbering: The “First Excited State” is $n=2$, NOT $n=1$. The ground state is $n=1$. The “Second Excited State” is $n=3$. Misreading this in a numerical will ruin the entire calculation.
Mistake 3
Ignoring the $Z^2$ factor: When working with ions like $\text{He}^+$ or $\text{Li}^{2+}$, students often use the $-13.6/n^2$ formula and forget to multiply by $Z^2$. For Helium ($Z=2$), the ground state energy is actually $-13.6 \times (2^2) = -54.4 \text{ eV}$.
Exam Tips
Tip 1
Number of Spectral Lines: If a sample of hydrogen gas is excited to the $n$-th energy level, the total possible number of distinct emission lines (as electrons cascade down to the ground state in various ways) is given by the formula: $\frac{n(n-1)}{2}$.
Tip 2
Energy Level Differences: Memorize the first four energy levels of Hydrogen to save calculation time: $E_1 = -13.6 \text{ eV}$, $E_2 = -3.4 \text{ eV}$, $E_3 = -1.51 \text{ eV}$, $E_4 = -0.85 \text{ eV}$. Notice that the energy gap drastically shrinks as $n$ increases!
Expected Exam Questions
SQ
Board Pattern Questions
Class 12 · Atoms · CBSE Exam1
The ground state energy of a hydrogen atom is $-13.6 \text{ eV}$. What are the kinetic and potential energies of the electron in this state? [2 marks]
Answer
$K = +13.6 \text{ eV}$, $U = -27.2 \text{ eV}$
📝
Explanation
Using the strict mathematical relation between energies in Bohr’s orbit:
Kinetic Energy $K = -E = -(-13.6 \text{ eV}) = +13.6 \text{ eV}$. (Kinetic energy is always positive).
Potential Energy $U = 2E = 2(-13.6 \text{ eV}) = -27.2 \text{ eV}$.
2
Calculate the shortest and longest wavelengths of the Balmer series of the hydrogen spectrum. (Given $R = 1.097 \times 10^7 \text{ m}^{-1}$) [3 marks]
Answer
Shortest $\approx 3646 \text{ \AA}$, Longest $\approx 6563 \text{ \AA}$
📝
Explanation
For the Balmer series, the final state is always $n_f = 2$.
1. Shortest Wavelength (Max Energy): Transition from $n_i = \infty$ to $n_f = 2$.
$\frac{1}{\lambda_S} = R \left( \frac{1}{2^2} – \frac{1}{\infty} \right) = \frac{R}{4}$
$\lambda_S = \frac{4}{R} = \frac{4}{1.097 \times 10^7} \approx 3.646 \times 10^{-7} \text{ m} = 3646 \text{ \AA}$.
2. Longest Wavelength (Min Energy): Transition from $n_i = 3$ to $n_f = 2$.
$\frac{1}{\lambda_L} = R \left( \frac{1}{2^2} – \frac{1}{3^2} \right) = R \left( \frac{1}{4} – \frac{1}{9} \right) = \frac{5R}{36}$
$\lambda_L = \frac{36}{5R} = \frac{36}{5 \times 1.097 \times 10^7} \approx 6.563 \times 10^{-7} \text{ m} = 6563 \text{ \AA}$.
3
A $5 \text{ MeV}$ alpha particle is projected towards a stationary gold nucleus ($Z = 79$). Calculate the distance of closest approach. [3 marks]
Answer
$r_0 = 4.55 \times 10^{-14} \text{ m}$
📝
Explanation
At the distance of closest approach ($r_0$), Kinetic Energy equals Electrostatic Potential Energy.
$K = 5 \text{ MeV} = 5 \times 10^6 \times 1.6 \times 10^{-19} \text{ J} = 8 \times 10^{-13} \text{ J}$.
$K = \frac{1}{4\pi\epsilon_0} \frac{(2e)(Ze)}{r_0}$
$r_0 = \frac{9 \times 10^9 \times 2 \times 79 \times (1.6 \times 10^{-19})^2}{8 \times 10^{-13}}$
$r_0 = \frac{9 \times 158 \times 2.56 \times 10^{-29}}{8 \times 10^{-13}} = \frac{3640.32 \times 10^{-29}}{8 \times 10^{-13}}$
$r_0 = 455.04 \times 10^{-16} \text{ m} = 4.55 \times 10^{-14} \text{ m}$.
Concept Map
Atoms connects to →
Quantum Physics
Dual Nature (De Broglie waves & Photons)
Nuclei (Nuclear dimensions)
Electromagnetism (Coulomb force in orbits)
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